$\sum\limits_{n=1}^\infty \frac{1}{\sqrt{1 + n^2}}$ Convergence

Question

I'm currently evaluating the sum

$$\sum\limits_{n=1}^\infty \frac{1}{\sqrt{1 + n^2}}$$

My strategy is writing the first few terms and seeing what sum I know they are similar to. If I do this I get:

$$n=1\rightarrow \frac{1}{\sqrt{2}} < 1$$ $$n=2\rightarrow \frac{1}{\sqrt{5}} < \frac{1}{2}$$ $$n=3\rightarrow \frac{1}{\sqrt{10}} < \frac{1}{3}$$ $$n=4\rightarrow \frac{1}{\sqrt{17}} < \frac{1}{4}$$

and so on.

My thought was (never mind an easier method, I just wondered if) to make this argument: you can make the difference between the members (n=1, n=2, ..) of this infinite sum and the harmonic series as small as you want, provided you pick $$n \geq N$$ for some $N$ which will depend on the difference of course. Formally I guess I want to say:

$$|a_n - b_n| < \epsilon$$ $$\forall n \geq N, \epsilon > 0$$ if we call $a_n$ the members of the harmonic series and $b_n$ the members of the infinite sum above.

Thus I can conclude that the sum diverges. Is there a mistake in my argument or have I (informally, of course) proved that this sum diverges?

Answer 1

Note that $$ 2n^2=n^2+n^2\geq1+n^2 $$ Then $$ \frac{1}{\sqrt{2}n}\leq\frac{1}{\sqrt{1+n^2}} $$

Answer 2

Use the limit comparison test. Note that $$ \frac{1}{\sqrt{1+n^2}}\sim \frac{1}{\sqrt{n^2}}=\frac{1}{n}\tag{1}. $$ Indeed, $$ \lim_{n\to\infty}\frac{1/\sqrt{1+n^2}}{1/n} =\lim_{n\to\infty}\frac{\sqrt{n^2}}{\sqrt{1+n^2}}= \lim_{n\to\infty}\frac{1}{\sqrt{1+n^{-2}}}=1. $$ By (1) since the harmonic series diverges, the series in question diverges as well.